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Differential equation y''-5y'+6y=2e^x+6x-5

For Cauchy problem:

y() =
y'() =
y''() =
y'''() =
y''''() =

The graph:

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The solution

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                          2                        
    d                    d                  x      
- 5*--(y(x)) + 6*y(x) + ---(y(x)) = -5 + 2*e  + 6*x
    dx                    2                        
                        dx                         
$$6 y{\left(x \right)} - 5 \frac{d}{d x} y{\left(x \right)} + \frac{d^{2}}{d x^{2}} y{\left(x \right)} = 6 x + 2 e^{x} - 5$$
6*y - 5*y' + y'' = 6*x + 2*exp(x) - 5
Detail solution
Given the equation:
$$6 y{\left(x \right)} - 5 \frac{d}{d x} y{\left(x \right)} + \frac{d^{2}}{d x^{2}} y{\left(x \right)} = 6 x + 2 e^{x} - 5$$
This differential equation has the form:
y'' + p*y' + q*y = s,

where
$$p = -5$$
$$q = 6$$
$$s = - 6 x - 2 e^{x} + 5$$
It is called linear inhomogeneous
second-order differential equation with constant coefficients.
The equation has an easy solution
We solve the corresponding homogeneous linear equation
y'' + p*y' + q*y = 0

First of all we should find the roots of the characteristic equation
$$q + \left(k^{2} + k p\right) = 0$$
In this case, the characteristic equation will be:
$$k^{2} - 5 k + 6 = 0$$
Detailed solution of the equation
- this is a simple quadratic equation
The roots of this equation:
$$k_{1} = 2$$
$$k_{2} = 3$$
As there are two roots of the characteristic equation,
and the roots are not complex, then
solving the correspondent differential equation looks as follows:
$$y{\left(x \right)} = C_{1} e^{k_{1} x} + C_{2} e^{k_{2} x}$$
$$y{\left(x \right)} = C_{1} e^{2 x} + C_{2} e^{3 x}$$

We get a solution for the correspondent homogeneous equation
Now we should solve the inhomogeneous equation
y'' + p*y' + q*y = s

Use variation of parameters method
Suppose that C1 and C2 - it is functions of x

The general solution is:
$$y{\left(x \right)} = \operatorname{C_{1}}{\left(x \right)} e^{2 x} + \operatorname{C_{2}}{\left(x \right)} e^{3 x}$$
where C1(x) and C2(x)
by the method of variation of parameters, we find the solution from the system:
$$\operatorname{y_{1}}{\left(x \right)} \frac{d}{d x} \operatorname{C_{1}}{\left(x \right)} + \operatorname{y_{2}}{\left(x \right)} \frac{d}{d x} \operatorname{C_{2}}{\left(x \right)} = 0$$
$$\frac{d}{d x} \operatorname{C_{1}}{\left(x \right)} \frac{d}{d x} \operatorname{y_{1}}{\left(x \right)} + \frac{d}{d x} \operatorname{C_{2}}{\left(x \right)} \frac{d}{d x} \operatorname{y_{2}}{\left(x \right)} = f{\left(x \right)}$$
where
y1(x) and y2(x) - linearly independent particular solutions of Linear Ordinary Differential Equations,
y1(x) = exp(2*x) (C1=1, C2=0),
y2(x) = exp(3*x) (C1=0, C2=1).
The free term f = - s, or
$$f{\left(x \right)} = 6 x + 2 e^{x} - 5$$
So, the system has the form:
$$e^{3 x} \frac{d}{d x} \operatorname{C_{2}}{\left(x \right)} + e^{2 x} \frac{d}{d x} \operatorname{C_{1}}{\left(x \right)} = 0$$
$$\frac{d}{d x} \operatorname{C_{1}}{\left(x \right)} \frac{d}{d x} e^{2 x} + \frac{d}{d x} \operatorname{C_{2}}{\left(x \right)} \frac{d}{d x} e^{3 x} = 6 x + 2 e^{x} - 5$$
or
$$e^{3 x} \frac{d}{d x} \operatorname{C_{2}}{\left(x \right)} + e^{2 x} \frac{d}{d x} \operatorname{C_{1}}{\left(x \right)} = 0$$
$$3 e^{3 x} \frac{d}{d x} \operatorname{C_{2}}{\left(x \right)} + 2 e^{2 x} \frac{d}{d x} \operatorname{C_{1}}{\left(x \right)} = 6 x + 2 e^{x} - 5$$
Solve the system:
$$\frac{d}{d x} \operatorname{C_{1}}{\left(x \right)} = \left(- 6 x - 2 e^{x} + 5\right) e^{- 2 x}$$
$$\frac{d}{d x} \operatorname{C_{2}}{\left(x \right)} = \left(6 x + 2 e^{x} - 5\right) e^{- 3 x}$$
- it is the simple differential equations, solve these equations
$$\operatorname{C_{1}}{\left(x \right)} = C_{3} + \int \left(- 6 x - 2 e^{x} + 5\right) e^{- 2 x}\, dx$$
$$\operatorname{C_{2}}{\left(x \right)} = C_{4} + \int \left(6 x + 2 e^{x} - 5\right) e^{- 3 x}\, dx$$
or
$$\operatorname{C_{1}}{\left(x \right)} = C_{3} + \left(3 x - 1\right) e^{- 2 x} + 2 e^{- x}$$
$$\operatorname{C_{2}}{\left(x \right)} = C_{4} + \left(1 - 2 x\right) e^{- 3 x} - e^{- 2 x}$$
Substitute found C1(x) and C2(x) to
$$y{\left(x \right)} = \operatorname{C_{1}}{\left(x \right)} e^{2 x} + \operatorname{C_{2}}{\left(x \right)} e^{3 x}$$
The final answer:
$$y{\left(x \right)} = C_{3} e^{2 x} + C_{4} e^{3 x} + x + e^{x}$$
where C3 and C4 is a constants
The answer [src]
               2*x       3*x    x
y(x) = x + C1*e    + C2*e    + e 
$$y{\left(x \right)} = C_{1} e^{2 x} + C_{2} e^{3 x} + x + e^{x}$$
The classification
nth linear constant coeff undetermined coefficients
nth linear constant coeff variation of parameters
nth linear constant coeff variation of parameters Integral
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